Đặt \(x^2-4x+10=a\ge6\)
\(\Leftrightarrow\frac{21}{a}-a+4=0\)
\(\Leftrightarrow-a^2+4a+21=0\) \(\Rightarrow\left[{}\begin{matrix}a=7\\a=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x^2-4x+10=7\Leftrightarrow x^2-4x+3=0\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Đặt \(x^2-4x=t\)
ta có pt : \(\frac{21}{t+10}-\left(t+6\right)=0\)
\(\Leftrightarrow21-\left(t+6\right)\left(t+10\right)=0\)
\(\Leftrightarrow21-t^2-10t-6t-60=0\)
\(\Leftrightarrow t^2+16t+39=0\)
\(\left[{}\begin{matrix}t=-3\\t=-13\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\\x^2-4x+13=0\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\\\left(x-2\right)^2+9>0\left(\text{vô nghiệm}\right)\end{matrix}\right.\)
vậy pt có nghiệm x = 1 và x = 3
